QUESTION IMAGE
Question
write the formulas and names for compounds of the following ions: chart with na+, al3+, n-3, br-, s2-
To solve for the chemical formulas, we use the criss - cross method where the charge of the cation becomes the subscript of the anion and vice - versa, and then we simplify if possible. We will go through each cation - anion pair:
1. For $\boldsymbol{Na^+}$ and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $N^{3 - }$ is $-3$. So, the formula is $Na_3N$.
Step 2: Name the compound
Sodium nitride.
2. For $\boldsymbol{Na^+}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $Br^-$ is $-1$. The formula is $NaBr$.
Step 2: Name the compound
Sodium bromide.
3. For $\boldsymbol{Na^+}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $S^{2 - }$ is $-2$. The formula is $Na_2S$.
Step 2: Name the compound
Sodium sulfide.
4. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $N^{3 - }$ is $-3$. We simplify the subscripts (divide by 3). The formula is $AlN$.
Step 2: Name the compound
Aluminum nitride.
5. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $Br^-$ is $-1$. The formula is $AlBr_3$.
Step 2: Name the compound
Aluminum bromide.
6. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $S^{2 - }$ is $-2$. The formula is $Al_2S_3$.
Step 2: Name the compound
Aluminum sulfide.
7. For $\boldsymbol{Mg^{2+}}$ (assuming the first cation is $Mg^{2+}$) and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $N^{3 - }$ is $-3$. The formula is $Mg_3N_2$.
Step 2: Name the compound
Magnesium nitride.
8. For $\boldsymbol{Mg^{2+}}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $Br^-$ is $-1$. The formula is $MgBr_2$.
Step 2: Name the compound
Magnesium bromide.
9. For $\boldsymbol{Mg^{2+}}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $S^{2 - }$ is $-2$. We simplify the subscripts (divide by 2). The formula is $MgS$.
Step 2: Name the compound
Magnesium sulfide.
If we tabulate the results:
| Cation \ Anion | $N^{3 - }$ | $Br^-$ | $S^{2 - }$ |
|---|---|---|---|
| $Al^{3+}$ | $AlN$ (Aluminum nitride) | $AlBr_3$ (Aluminum bromide) | $Al_2S_3$ (Aluminum sulfide) |
| $Mg^{2+}$ | $Mg_3N_2$ (Magnesium nitride) | $MgBr_2$ (Magnesium bromide) | $MgS$ (Magnesium sulfide) |
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To solve for the chemical formulas, we use the criss - cross method where the charge of the cation becomes the subscript of the anion and vice - versa, and then we simplify if possible. We will go through each cation - anion pair:
1. For $\boldsymbol{Na^+}$ and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $N^{3 - }$ is $-3$. So, the formula is $Na_3N$.
Step 2: Name the compound
Sodium nitride.
2. For $\boldsymbol{Na^+}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $Br^-$ is $-1$. The formula is $NaBr$.
Step 2: Name the compound
Sodium bromide.
3. For $\boldsymbol{Na^+}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Na^+$ is $+ 1$ and the charge of $S^{2 - }$ is $-2$. The formula is $Na_2S$.
Step 2: Name the compound
Sodium sulfide.
4. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $N^{3 - }$ is $-3$. We simplify the subscripts (divide by 3). The formula is $AlN$.
Step 2: Name the compound
Aluminum nitride.
5. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $Br^-$ is $-1$. The formula is $AlBr_3$.
Step 2: Name the compound
Aluminum bromide.
6. For $\boldsymbol{Al^{3+}}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Al^{3+}$ is $+ 3$ and the charge of $S^{2 - }$ is $-2$. The formula is $Al_2S_3$.
Step 2: Name the compound
Aluminum sulfide.
7. For $\boldsymbol{Mg^{2+}}$ (assuming the first cation is $Mg^{2+}$) and $\boldsymbol{N^{3 - }}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $N^{3 - }$ is $-3$. The formula is $Mg_3N_2$.
Step 2: Name the compound
Magnesium nitride.
8. For $\boldsymbol{Mg^{2+}}$ and $\boldsymbol{Br^-}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $Br^-$ is $-1$. The formula is $MgBr_2$.
Step 2: Name the compound
Magnesium bromide.
9. For $\boldsymbol{Mg^{2+}}$ and $\boldsymbol{S^{2 - }}$
Step 1: Apply criss - cross method
The charge of $Mg^{2+}$ is $+ 2$ and the charge of $S^{2 - }$ is $-2$. We simplify the subscripts (divide by 2). The formula is $MgS$.
Step 2: Name the compound
Magnesium sulfide.
If we tabulate the results:
| Cation \ Anion | $N^{3 - }$ | $Br^-$ | $S^{2 - }$ |
|---|---|---|---|
| $Al^{3+}$ | $AlN$ (Aluminum nitride) | $AlBr_3$ (Aluminum bromide) | $Al_2S_3$ (Aluminum sulfide) |
| $Mg^{2+}$ | $Mg_3N_2$ (Magnesium nitride) | $MgBr_2$ (Magnesium bromide) | $MgS$ (Magnesium sulfide) |